Since the function is continuous on [-2,1] there is a maximum and a minimum. The extrema can occur at critical points or the endpoints of the interval. Critical points occur where the first derivative is zero or fails to exist.

First we compute the first derivative:

Using the quotient rule we get `f'(x)=d/(dx)[(2x)/((1+5x^2)^2)]`

`=((1+5x^2)^2(2)-(2x)(2)(1+5x^2)(10x))/((1+5x^2)^4)`

`=((1+5x^2)(2(1+5x^2)-40x^2))/((1+5x^2)^4)`

`=(2-30x^2)/((1+5x^2)^3)`

`f'(x)=0==>2-30x^2=0==>x=+-sqrt(1/15)`

We evaluate f at `-2,-sqrt(1/15),sqrt(1/15),` and 1.

`f(-2)~~-.0091`

`f(-sqrt(1/15))~~-.2905`

`f(sqrt(1/15))~~.2905`

`f(1)=.0bar(5)`

----------------------------------------------------------------

**The maximum on the interval occurs at `x=sqrt(1/15)` with a value of `f(sqrt(1/15))~~.2905` .** ** `-sqrt(1/15)~~-.2582`

**The minimum on the interval occurs at `x=-sqrt(1/15)` with a value of `f(-sqrt(1/15))~~-.2905` ** ** `sqrt(1/15)~~.2582`

----------------------------------------------------------------

The graph: